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Chapter 11. Poisson processes.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
c
2008.
Miguel A. Arcones. All rights reserved.
Extract from:
”Arcones’ Manual for SOA Exam MLC. Fall 2009 Edition”,
available at http://www.actexmadriver.com/
1/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Exponential distribution
Definition 1
The gamma function is defined by
Z ∞
Γ(α) =
t α−1 e −t dt, α > 0.
0
2/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Exponential distribution
Definition 1
The gamma function is defined by
Z ∞
Γ(α) =
t α−1 e −t dt, α > 0.
0
For example,
Z
Γ(1) =
∞
t 1−1 e −t dt =
0
Z
∞
e −t dt = 1.
0
3/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Exponential distribution
Definition 1
The gamma function is defined by
Z ∞
Γ(α) =
t α−1 e −t dt, α > 0.
0
For example,
Z
Γ(1) =
∞
t 1−1 e −t dt =
0
Z
∞
e −t dt = 1.
0
Γ(α) < ∞, for α > 0, because
Z ∞
Z 1
Z ∞
α−1 −t
α−1 −t
t
e dt =
t
e dt +
t α−1 e −t dt
0
0
1
∞
Z 1
Z ∞
t α 1
1
α−1
−t
−t ≤
t
dt +
e dt =
+e
= + e −1 .
α 0
α
0
1
1
4/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 1
The gamma function satisfies the following properties:
(i) For each α > 1, Γ(α) = (α − 1)Γ(α − 1).
(ii) For each integer n ≥ 1, Γ(n) = (n − 1)!.
√
(iii) Γ(1/2) = π.
5/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 1
The gamma function satisfies the following properties:
(i) For each α > 1, Γ(α) = (α − 1)Γ(α − 1).
(ii) For each integer n ≥ 1, Γ(n) = (n − 1)!.
√
(iii) Γ(1/2) = π.
Proof:
(i) For each α > 1, by an integration by parts
Z ∞
Z ∞
α−1 −t
Γ(α) =
t
e dt =
t α−1 d(−e −t ) dt
0
0
∞ Z ∞
α−1
−t =t
(−e ) −
(−e −t )d(t α−1 )
0
0
Z ∞
=
(α − 1)t α−2 e −t dt = (α − 1)Γ(α − 1).
0
6/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 1
The gamma function satisfies the following properties:
(i) For each α > 1, Γ(α) = (α − 1)Γ(α − 1).
(ii) For each integer n ≥ 1, Γ(n) = (n − 1)!.
√
(iii) Γ(1/2) = π.
Proof:
(ii) We use induction to prove that for each integer n ≥ 1, Γ(n) =
(n − 1)!. We have that Γ(1) = 1. The case n − 1 implies the case
n, because if Γ(n − 1) = (n − 2)!, then Γ(n) = (n − 1)Γ(n − 1) =
(n − 1)(n − 2)! = (n − 1)!.
7/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 1
The gamma function satisfies the following properties:
(i) For each α > 1, Γ(α) = (α − 1)Γ(α − 1).
(ii) For each integer n ≥ 1, Γ(n) = (n − 1)!.
√
(iii) Γ(1/2) = π.
Proof:
(iii) By the change of variables t = x 2 /2, (dt = x dx),
√ R ∞ − x2
R∞
R ∞ q 2 − x2
2 x dx =
Γ(1/2) = 0 t −1/2 e −t dt = 0
e
2 0 e 2 dx
2
x
√ R
√
2
√
√
x
∞
= 22 −∞ e − 2 dx = 22 2π = π
8/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
In particular, we have that
Z ∞
x n e −x dx = Γ(n + 1) = n!, n ≥ 1.
(1)
0
By the change of variables y = xθ ,
Z ∞
Z ∞
α−1 −x/θ
x
e
dx =
y α−1 θα e −y dy = θα Γ(α), α, θ > 0. (2)
0
0
9/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 1
Find:
R∞
(i) R0 x 3 e −x dx.
∞
(ii) R0 x 12 e −x dx.
∞
(iii) 0 x 23 e −2x dx.
R ∞ 24 −x/3
(iv) R 0 x e
dx.
∞ 3.5 −x
(v) 0 x e dx.
10/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 1
Find:
R∞
(i) R0 x 3 e −x dx.
∞
(ii) R0 x 12 e −x dx.
∞
(iii) 0 x 23 e −2x dx.
R ∞ 24 −x/3
(iv) R 0 x e
dx.
∞ 3.5 −x
(v) 0 x e R dx.
∞
Solution: (i) 0 x 3 e −x dx = 3! = 6.
11/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 1
Find:
R∞
(i) R0 x 3 e −x dx.
∞
(ii) R0 x 12 e −x dx.
∞
(iii) 0 x 23 e −2x dx.
R ∞ 24 −x/3
(iv) R 0 x e
dx.
∞ 3.5 −x
(v) 0 x e R dx.
∞ 3 −x
Solution:
R ∞ 12(i)−x 0 x e dx = 3! = 6.
(ii) 0 x e dx = (12)! = 479001600.
12/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 1
Find:
R∞
(i) R0 x 3 e −x dx.
∞
(ii) R0 x 12 e −x dx.
∞
(iii) 0 x 23 e −2x dx.
R ∞ 24 −x/3
(iv) R 0 x e
dx.
∞ 3.5 −x
(v) 0 x e R dx.
∞ 3 −x
Solution:
R ∞ 12(i)−x 0 x e dx = 3! = 6.
(ii) R0 x e dx = (12)! = 479001600.
∞
(iii) 0 x 23 e −2x dx = (23)! 2124 = 1540900274448694.
13/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 1
Find:
R∞
(i) R0 x 3 e −x dx.
∞
(ii) R0 x 12 e −x dx.
∞
(iii) 0 x 23 e −2x dx.
R ∞ 24 −x/3
(iv) R 0 x e
dx.
∞ 3.5 −x
(v) 0 x e R dx.
∞ 3 −x
Solution:
R ∞ 12(i)−x 0 x e dx = 3! = 6.
(ii) R0 x e dx = (12)! = 479001600.
∞
(iii) 0 x 23 e −2x dx = (23)! 2124 = 1540900274448694.
R ∞ 24 −x/3
(iv) 0 x e
dx = (24)!325 = (5.256988635)(10)035 .
14/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 1
Find:
R∞
(i) R0 x 3 e −x dx.
∞
(ii) R0 x 12 e −x dx.
∞
(iii) 0 x 23 e −2x dx.
R ∞ 24 −x/3
(iv) R 0 x e
dx.
∞ 3.5 −x
(v) 0 x e R dx.
∞ 3 −x
Solution:
R ∞ 12(i)−x 0 x e dx = 3! = 6.
(ii) R0 x e dx = (12)! = 479001600.
∞
(iii) 0 x 23 e −2x dx = (23)! 2124 = 1540900274448694.
R ∞ 24 −x/3
(iv) R 0 x e
dx = (24)!325 = (5.256988635)(10)035 .
∞ 3.5 −x
(v) 0 x e dx = Γ(3.5) = (2.5)Γ(2.5) = (2.5)(1.5)Γ(1.5) =
√
(2.5)(1.5)(0.5) π.
15/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 2
For each integer n ≥ 1,
Z
n
X xj
x n −x
e dx = −e −x
+ c.
n!
j!
j=0
16/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Proof.
We proceed by induction. If n = 1, we have that
Z
Z
Z
xe −x dx = xd(−e −x ) = x(−e −x ) − (−e −x ) dx
=x(−e −x ) − e −x + c = −e −x (1 + x) + c.
Assume that the claim holds for the case n − 1. Then,
n
Z n
Z n
Z
x −x
x
xn
x
e dx =
d(−e −x ) =
(−e −x ) − (−e −x ) d
n!
n!
n!
n!
Z
n−1
X
xn
x n−1
xn
xj
= − e −x + e −x
dx = − e −x −
e −x + c
n!
(n − 1)!
n!
j!
j=0
=−
n
X
j=0
e −x
xj
+ c.
j!
17/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Definition 2
A r.v. X is said to have an exponential distribution with
parameter λ > 0, if the density of X is given by
( −x
e λ
if x ≥ 0
λ
f (x) =
0
if x < 0
We denote this by X ∼ Exponential(λ).
18/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Definition 2
A r.v. X is said to have an exponential distribution with
parameter λ > 0, if the density of X is given by
( −x
e λ
if x ≥ 0
λ
f (x) =
0
if x < 0
We denote this by X ∼ Exponential(λ).
The above function f defines a legitime density because it is
nonnegative and
Z
∞
Z
f (t) dt =
−∞
0
∞
∞
t
e− λ
− λt dt = −e
= 1.
λ
0
19/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 3
Let X be a r.v. with an exponential distribution with parameter
λ > 0, then
E [X ] = λ, Var(X ) = λ2 , E [X k ] = λk k!, M(t) =
1
, if t < λ−1 .
1 − λt
20/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 3
Let X be a r.v. with an exponential distribution with parameter
λ > 0, then
E [X ] = λ, Var(X ) = λ2 , E [X k ] = λk k!, M(t) =
1
, if t < λ−1 .
1 − λt
Proof: Using that
Z
∞
x α−1 e −x/θ dx = θα Γ(α),
0
we get that
k
Z
E [X ] =
∞
x
0
ke
− λx
λ
dx =
1
Γ(k + 1)λk+1 = k!λk
λ
In particular, E [X ] = λ and E [X 2 ] = 2λ2 . Hence,
Var(X ) = E [X 2 ] − (E [X ])2 = 2λ2 − λ2 = λ2 .
21/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
We have that for t < λ−1 ,
Z ∞
Z
− λx
1 ∞ −x( 1−λt )
tX
tx e
λ
dx
M(t) = E [e ] =
e
dx =
e
λ
λ 0
0
1 λ
1
=
=
.
λ 1 − λt
1 − λt
22/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 2
The time that it takes for a machine to fail is exponential with
mean 1000 days. Find the median of the failure time.
23/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 2
The time that it takes for a machine to fail is exponential with
mean 1000 days. Find the median of the failure time.
Solution: Let m be the median of the failure time. The failure
x
1
time has density 1000
e − 1000 . Hence,
∞
Z ∞
x
m
1 − x
1
−
=
e 1000 dx = −e 1000 = e − 1000
2
m 1000
m
and m = 1000 log(2) = 693.1471806.
24/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
The cumulative distribution function of an exponential distribution
with mean λ > 0 is
Z x
Z x −t
x
e λ
dt = 1−e − λ , x ≥ 0.
F (x) = P{X ≤ x} =
f (t) dt =
−∞
−∞ λ
25/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
The exponential distribution satisfies that for each s, t ≥ 0,
P{X > s + t|X > t} = P{X > s}.
This is property is called the memoryless property of the
exponential distribution. Notice that
s+t
s
e− λ
P{X > s + t}
=
= e − λ = P{X > s}.
P{X > s+t|X > t} =
t
−
P{X > t}
e λ
26/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 3
The life time T of a radioactive substance is an exponential
distributed random variable with mean 1.5 years.
(i) What is the probability that the lifetime of a sample of this
substance exceeds 2 years.
(ii) What is the probability that a sample of radioactive substance
is present 14 years given that is duration exceeds 12 years.
27/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 3
The life time T of a radioactive substance is an exponential
distributed random variable with mean 1.5 years.
(i) What is the probability that the lifetime of a sample of this
substance exceeds 2 years.
(ii) What is the probability that a sample of radioactive substance
is present 14 years given that is duration exceeds 12 years.
2t
Solution: (i) T has density f (t) = 32 e − 3 , t ≥ 0. So,
∞
Z ∞
2t 4
2 − 2t
−
P{T ≥ 2} =
e 3 dt = −e 3 = e − 3 ,
3
2
2
28/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 3
The life time T of a radioactive substance is an exponential
distributed random variable with mean 1.5 years.
(i) What is the probability that the lifetime of a sample of this
substance exceeds 2 years.
(ii) What is the probability that a sample of radioactive substance
is present 14 years given that is duration exceeds 12 years.
2t
Solution: (i) T has density f (t) = 32 e − 3 , t ≥ 0. So,
∞
Z ∞
2t 4
2 − 2t
−
P{T ≥ 2} =
e 3 dt = −e 3 = e − 3 ,
3
2
2
(ii)
4
P{T ≥ 14|T ≥ 12} = P{T ≥ 2} = e − 3 .
29/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Definition 3
X has a gamma distribution with parameters α > 0 and β > 0, if
the density of X is

x
 x α−1 e − β
if x ≥ 0
α Γ(α)
β
f (x) =
0
if x < 0
We denote this by X ∼ Gamma(α, β).
30/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Definition 3
X has a gamma distribution with parameters α > 0 and β > 0, if
the density of X is

x
 x α−1 e − β
if x ≥ 0
α Γ(α)
β
f (x) =
0
if x < 0
We denote this by X ∼ Gamma(α, β).
The above function f defines a legitime density because by (2)
Z
∞
Z
f (x) dx =
−∞
0
∞
−x
x α−1 e β
dx = 1.
Γ(α)β α
31/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Definition 3
X has a gamma distribution with parameters α > 0 and β > 0, if
the density of X is

x
 x α−1 e − β
if x ≥ 0
α Γ(α)
β
f (x) =
0
if x < 0
We denote this by X ∼ Gamma(α, β).
The above function f defines a legitime density because by (2)
Z
∞
Z
f (x) dx =
−∞
0
∞
−x
x α−1 e β
dx = 1.
Γ(α)β α
A gamma distribution with parameter α = 1 is an exponential
distribution.
32/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 4
If X has a gamma distribution with parameters α and β, then
E [X ] = αβ, Var(X ) = αβ 2 , E [X k ] =
M(t) =
Γ(α + k)β k
,
Γ(α)
1
1
, if t < .
(1 − βt)α
β
33/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 4
If X has a gamma distribution with parameters α and β, then
E [X ] = αβ, Var(X ) = αβ 2 , E [X k ] =
M(t) =
Γ(α + k)β k
,
Γ(α)
1
1
, if t < .
(1 − βt)α
β
Proof: Using that
Z
∞
x α−1 e −x/θ dx = θα Γ(α),
0
Z
∞
x
α−1 e − β
1
dx =
E [X ] =
x
α
Γ(α)β
Γ(α)β α
0
1
Γ(α + k)β α
k+α
=
Γ(k
+
α)β
=
.
Γ(α)β α
Γ(α)
k
kx
Z
∞
x k+α−1 e
− βx
dx
0
34/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 4
If X has a gamma distribution with parameters α and β, then
E [X ] = αβ, Var(X ) = αβ 2 , E [X k ] =
M(t) =
1
1
, if t < .
(1 − βt)α
β
Proof: Using that E [X k ] =
E [X ] =
Γ(α + k)β k
,
Γ(α)
Γ(α+k)β α
,
Γ(α)
Γ(α + 1)β 1
Γ(α + 2)β 2
= αβ, E [X 2 ] =
= (α + 1)αβ 2 .
Γ(α)
Γ(α)
Hence,
Var(X ) = E [X 2 ] − (E [X ])2 = (α + 1)αβ 2 − (αβ)2 = αβ 2 .
35/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 4
If X has a gamma distribution with parameters α and β, then
E [X ] = αβ, Var(X ) = αβ 2 , E [X k ] =
M(t) =
Γ(α + k)β k
,
Γ(α)
1
1
, if t < .
(1 − βt)α
β
Proof: We have that for t < β1 ,
−x
∞
x α−1 e β
M(t) = E [e ] =
e
dx
Γ(α)β α
0
α
“
”
Z ∞
1−βt
1
β
1
α−1 −x
β
=
Γ(α)
x
e
dx =
Γ(α)β α 0
Γ(α)β α 1 − βt
1
=
.
(1 − βt)α
tX
Z
tx
36/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 4
Find the mean and the variance of a random variable X with
density
x
x 3e − 3
f (x) =
, x ≥ 0.
486
37/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 4
Find the mean and the variance of a random variable X with
density
x
x 3e − 3
f (x) =
, x ≥ 0.
486
Solution: We have that α = 4, and β = 3. Hence,
E [X ] = αβ = 12 and Var(X ) = αβ 2 = 36.
38/58
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2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 5
Let X ∼ Gamma(α1 , β) and Y ∼ Gamma(α2 , β). Suppose that
X and Y are independent. Then, X + Y ∼ Gamma(α1 + α2 , β)
39/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 5
Let X ∼ Gamma(α1 , β) and Y ∼ Gamma(α2 , β). Suppose that
X and Y are independent. Then, X + Y ∼ Gamma(α1 + α2 , β)
Proof.
The moment generating function of X and Y are
1
1
MX (t) = (1−βt)
α1 and MY (t) = (1−βt)α2 , respectively. Since X
and Y are independent r.v.’s, the moment generating function of
X + Y is
MX +Y (t) = MX (t)MY (t) =
=
1
1
α
(1 − βt) 1 (1 − βt)α2
1
,
(1 − βt)α1 +α2
which is the moment generating function of a gamma distribution
with parameters α1 + α2 and β. So, X + Y has a gamma
distribution with parameters α1 + α2 and β.
40/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Notice that by induction, the previous theorem implies that if
X1 , . . . , Xn are independent
r.v.’s and P
Xi ∼ Gamma(αi , β),
Pn
1 ≤ i ≤ n, then, i=1 Xi ∼ Gamma( ni=1 αi , β).
41/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 6
Let X1 , . . . , Xn be independent identically distributed
Pn r.v.’s with
exponential distribution with parameter λ. Then, i=1 Xi has a
gamma distribution with parameters n and λ.
42/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 6
Let X1 , . . . , Xn be independent identically distributed
Pn r.v.’s with
exponential distribution with parameter λ. Then, i=1 Xi has a
gamma distribution with parameters n and λ.
Proof.
Every Xi has Gamma(1, λ) distribution. By the previous theorem,
P
n
i=1 Xi has a Gamma(n, λ) distribution.
43/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 5
Suppose that you arrive at a single–teller office of the Department
of Motor Vehicles to find three customers waiting in line and one
being served. If the services times are all exponential with rate 2
minutes, calculate the probability that you will have to wait in line
more than 10 minutes before being served.
44/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 5
Suppose that you arrive at a single–teller office of the Department
of Motor Vehicles to find three customers waiting in line and one
being served. If the services times are all exponential with rate 2
minutes, calculate the probability that you will have to wait in line
more than 10 minutes before being served.
Solution: By the memory–less property of the exponential distribution, the remaining serving time for the customer which
P is served is
also exponential. Hence, your waiting time is Y = 4j=1 Xj , where
{Xj }4j=1 are independent identically distributed r.v.’s with an exponential distribution with mean 2. By Theorem 6, Y has a gamma
distribution with parameters 4 and 2. Hence, the density of Y is
y
y
y 3e − 2
y 3e − 2
=
.
fY (y ) = 4
2 (3!)
96
45/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 5
Suppose that you arrive at a single–teller office of the Department
of Motor Vehicles to find three customers waiting in line and one
being served. If the services times are all exponential with rate 2
minutes, calculate the probability that you will have to wait in line
more than 10 minutes before being served.
Solution: By the change of variable y2 = z,
y
Z ∞ 3 −z
y 3e − 2
z e
P{Y ≥ 10} =
dy =
dz
96
6
10
5
∞ 3
3
z
z2
5
52
=−
+
+ z + 1 e −z =
+
+ 5 + 1 e −5
6
2
6
2
5
Z
∞
=0.2650259153.
46/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Suppose that a system has n parts. The system functions works
only if all n parts work. Let Xi is the lifetime of the i–th part of
the system. Suppose that X1 , . . . , Xn be independent r.v.’s and
that Xi has an exponential distribution with mean θi . Let Y be the
lifetime of the system. Then, Y = min(X1 , . . . , Xn ). Then,
P{Y > t} = P{min(X1 , . . . , Xn ) > t} = P{∩ni=1 {Xi > t}}
n
n
P
Y
Y
−t
−t ni=1 θ1
i .
=
P[Xi > t} =
e θi = e
i=1
i=1
So, Y has an exponential distribution with mean
Pn 1
1
i=1 θi
.
47/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 6
A system consists of 4 components. The lifetime of the four
component are independent random variables with an exponential
distribution and respective means 2,3,4, 10. The system will work
only if all four components work. Find the expected lifetime of the
system.
48/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 6
A system consists of 4 components. The lifetime of the four
component are independent random variables with an exponential
distribution and respective means 2,3,4, 10. The system will work
only if all four components work. Find the expected lifetime of the
system.
Solution: Let Xi , 1 ≤ i ≤ 4, are the lifetime of the components.
Let T = min1≤i≤4 Xi be the lifetime of the system. We have that
1
E [T ] = Pn
1
i=1 θi
=
1
2
+
1
3
1
+ 14 +
1
10
= 0.8450704225.
49/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 7
A system consists of 4 components. The lifetime of the four
component are independent random variables with an exponential
distribution and respective means 2,3,4, 10. The system will work
only if all four components work. Find the expected lifetime of the
system.
50/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 7
A system consists of 4 components. The lifetime of the four
component are independent random variables with an exponential
distribution and respective means 2,3,4, 10. The system will work
only if all four components work. Find the expected lifetime of the
system.
Solution: Let Xi , 1 ≤ i ≤ 4, are the lifetime of the components.
Let T = min1≤i≤4 Xi be the lifetime of the system. We have that
1
E [T ] = Pn
1
i=1 θi
=
1
2
+
1
3
1
+ 14 +
1
10
= 0.8450704225.
51/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Theorem 7
(i) Let X and Y be two independent r.v.’s such that X has an
exponential distribution with mean θ1 and Y has an exponential
distribution with mean θ2 . Then,
P{X < Y } =
1
θ1
1
θ1
+
1
θ2
.
(ii) Let X1 , . . . , Xn be independent r.v.’s such that Xi has an
exponential distribution with mean θi . Then,
P{Xi = min Xj } =
1≤j≤n
1
θi
Pn 1
j=1 θj
.
52/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Proof: (i) The joint density of X and Y is
fX ,Y (x, y ) =
e
− θx − θy
1
2
θ1 θ2
, x, y > 0.
So,
P{X < Y } =
R∞R∞
0
x
e
y
− x −
θ 1 θ2
θ1 θ2
dy dx =
R∞
0
e
− x − x
θ 1 θ2
θ1
dx =
1
θ1
1
+ θ1
θ1
2
53/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Proof: (ii)
P{Xi = min Xj } = P{Xi <
1≤j≤n
min
1≤j≤n,j6=i
Xj } =
1
θi
Pn 1
j=1 θj
,
because min1≤j≤n,j6=i Xj has an exponential distribution with mean
−1
P
n
1
and Xi and min1≤j≤n,j6=i Xj are independent.
1≤j≤n,j6=i θj
54/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Example 8
A factory has two electricity generators. The smaller of the two
generators has expected duration before failure of 20 days. The
other generator has an expected duration of 15 days. The amount
of time which each generator lasts before failing has an exponential
distribution. The duration before failure of the two generators are
independent r.v.’s.
(i) Calculate the mean of the time until one of the two generators
fails.
(ii) Calculate the mean of the time until both generators breaks
down.
(iii) Calculate the probability that the smaller generators fails
before the other one.
55/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Solution: (i) Let X be the lifetime of the smaller generator. X has
an exponential distribution with mean 20. Let Y be the lifetime of
the larger generator. Y has an exponential distribution with mean
15. The mean of the time until one of the two generators fails is
E [min(X , Y )] =
1
20
1
+
1
15
=
60
.
7
56/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Solution: (i) Let X be the lifetime of the smaller generator. X has
an exponential distribution with mean 20. Let Y be the lifetime of
the larger generator. Y has an exponential distribution with mean
15. The mean of the time until one of the two generators fails is
E [min(X , Y )] =
1
20
1
+
1
15
=
60
.
7
(ii) E [max(X , Y )] = E [X + Y − min(X , Y )] = 20 + 15 − 60
7 =
185
7 .
57/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.
Chapter 11. Poisson processes.
Section 11.1. Exponential and gamma distributions.
Solution: (i) Let X be the lifetime of the smaller generator. X has
an exponential distribution with mean 20. Let Y be the lifetime of
the larger generator. Y has an exponential distribution with mean
15. The mean of the time until one of the two generators fails is
E [min(X , Y )] =
1
20
1
+
1
15
=
60
.
7
(ii) E [max(X , Y )] = E [X + Y − min(X , Y )] = 20 + 15 − 60
7 =
(iii) P{X < Y } =
1
20
1
1
+
20
15
185
7 .
= 73 .
58/58
c
2008.
Miguel A. Arcones. All rights reserved.
Manual for SOA Exam MLC.